Question

Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, how many electrons were transferred from one sphere to the other during rubbing?

Answer 1

Explanation:

Step 1:

Given data,

$F=0.1 \mathrm{N}$

$r=10^{-2} \mathrm{m}$

Where  

F is attractive force between spheres,

r is spheres separation

Step 2:

Electrostatic force by Coulomb's Law,        

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$  

Step 3:

Let the electrons no. being moved from one sphere to another be n. Then

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{(n q)^{2}}{r^{2}}$

$0.1=9 \times 10^{9} \times \frac{\left(1.6 \times 10^{-19}\right)^{2} \times n^{2}}{10^{-4}}$

$0.1 \times 10^{-4}=9 \times 10^{9} \times\left(2.56 \times 10^{-38}\right) \times \mathrm{n}^{2}$

$10^{-5}=2.304 \times 10^{-28} \times n^{2}$

$\mathrm{n}^{2}=\frac{10^{-5}}{2.304 \times 10^{-28}}$

$\mathrm{n}^{2}=4.34 \times 10^{22}$

$\mathrm{n}=\sqrt{4.34 \times 10^{22}}$

$\mathrm{n}=2.08 \times 10^{11}$

During process of rubbing, electrons were moved from one sphere to another